HDU 1540 Tunnel Warfare 【线段树】

最新推荐文章于 2023-04-09 15:28:08 发布

Head_Hard

最新推荐文章于 2023-04-09 15:28:08 发布

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分类专栏：线段树

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本文介绍了一个基于线段树的数据结构实现的抗日地道战模拟系统。该系统能够处理村庄被破坏后的连接状态查询，并能在村庄重建后更新其连接状态。通过对具体案例的解析，深入探讨了线段树的应用及其实现细节。

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Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10962 Accepted Submission(s): 4305

Problem Description

During the War of Resistance Against Japan, tunnel warfare was carried out extensively in the vast areas of north China Plain. Generally speaking, villages connected by tunnels lay in a line. Except the two at the ends, every village was directly connected with two neighboring ones.

Frequently the invaders launched attack on some of the villages and destroyed the parts of tunnels in them. The Eighth Route Army commanders requested the latest connection state of the tunnels and villages. If some villages are severely isolated, restoration of connection must be done immediately!

Input

The first line of the input contains two positive integers n and m (n, m ≤ 50,000) indicating the number of villages and events. Each of the next m lines describes an event.

There are three different events described in different format shown below:

D x: The x-th village was destroyed.

Q x: The Army commands requested the number of villages that x-th village was directly or indirectly connected with including itself.

R: The village destroyed last was rebuilt.

Output

Output the answer to each of the Army commanders’ request in order on a separate line.

Sample Input

7 9
D 3
D 6
D 5
Q 4
Q 5
R
Q 4
R
Q 4

Sample Output

有很多细节的一个题，磨了足足三天，ac之后对基础线段树又加深了见解， 还是太菜了，继续加油吧

#include<stack>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define mms(x, y) memset(x, y, sizeof x)
#define MAX 50005
#define INF 0x3f3f3f3f
int mx[MAX << 2], mn[MAX << 2];
int n, m;
stack <int> s;
void PushUp(int rt)
{
    mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
    mn[rt] = min(mn[rt << 1], mn[rt << 1 | 1]);
}
void BuildTree(int rt, int l, int r)
{
    if(l == r)
    {
        mx[rt] = 0;
        mn[rt] = INF;
        return;
    }
    int mid = (l + r) >> 1;
    BuildTree(lson);
    BuildTree(rson);
    PushUp(rt);
}
void Update(int rt, int l, int r, int i, int val)
{
    if(l == r)
    {
        if(val == 0)
        {
            mx[rt] = 0;
            mn[rt] = INF;
            return;
        }
        else
        {
            mx[rt] = mn[rt] = val;
            return;
        }
    }
    int mid = (l + r) >> 1;
    if(i <= mid)
        Update(lson, i, val);
    if(i > mid)
        Update(rson, i, val);
    PushUp(rt);
}
int QueryA(int rt, int l, int r, int x, int y)
{
    if(l >= x && y >= r)
        return mx[rt];
    int mid = (l + r) >> 1, ans = 0;
    if(mid >= x)
        ans = max(QueryA(lson, x, y), ans);
    if(y > mid)
        ans = max(QueryA(rson, x, y), ans);
    return ans;
}
int QueryI(int rt, int l, int r, int x, int y)
{
    if(l >= x && y >= r)
        return mn[rt];
    int mid = (l + r) >> 1, ans = INF;
    if(mid >= x)
        ans = min(QueryI(lson, x, y), ans);
    if(y > mid)
        ans = min(QueryI(rson, x, y), ans);
    return ans;
}
void init()
{
    mms(mx, 0);
    mms(mn, 0);
    BuildTree(1, 1, n);
    while(!s.empty())
        s.pop();
}
int main()
{
    while(scanf("%d%d", &n, &m) != EOF)
    {
        char ss[5];
        int a;
        init();
        while(m--)
        {
            scanf("%s", ss);
            if(ss[0] == 'D')
            {
                scanf("%d", &a);
                s.push(a);
                Update(1, 1, n, a, a);
            }
            if(ss[0] == 'R')
            {
                a = s.top();
                s.pop();
                Update(1, 1, n, a, 0);
            }
            if(ss[0] == 'Q')
            {
                scanf("%d", &a);
                if(QueryA(1, 1, n, 1, a) == a && QueryI(1, 1, n, a, n) == a)
                    printf("0\n");
                else
                {
                    int l = a - QueryA(1, 1, n, 1, a);
                    int r;
                    if(QueryI(1, 1, n, a, n) == INF)
                        r = n - a + 1;
                    else
                        r = QueryI(1, 1, n, a, n) - a;
                    printf("%d\n", l + r - 1);
                }
            }
        }
    }
    return 0;
}